# What spaces can be obtained from $\mathbb{R}^{n}$ by taking quotient spaces and subspaces?

Is there a good characterization of the smallest collection of topological spaces which contains $\mathbb{R}^{n}$ for each $n$, and is closed under taking subspaces and quotient spaces?

A bit of motivation: A friend of mine asked me to give an argument why the definition of a topological space is "right" or "natural", considered perhaps as a generalization of manifolds or cell complexes. While trying to answer him, I briefly wondered whether the collection of topological spaces is the closure of $\{ \mathbb{R}^{n} \}_{n \geq 0}$ under certain operations, say taking subspaces and quotient spaces. I quickly realized that this is false in general, though (there are counterexamples which have very large cardinality or don't satisfy first or second countability).

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Your motivation question seems much more interesting! Related questions have been discussed on MO before, e.g. mathoverflow.net/questions/19152/… . The answers that most convinced me were the ones involving logic and computability. They suggest that the definition of a topology is useful because it is absurdly general, hence general enough to include nice things. But it is not necessarily geometrically natural. I think Grothendieck once expressed an opinion that the definition is "wrong" e.g. for homotopy theory? –  Qiaochu Yuan Mar 8 '11 at 10:40
If you used quotients and topological sums, you would get sequential spaces. Using subspaces, quotiens and sums, you would get subsequential spaces. (S. P. Franklin, M. Rajagopalan: On subsequential spaces, Topology. and its Applications 35 (1990), 1–19) Your class will definiely be a subclass of the class of subsequential spaces. I am not sure about the precise characterization. –  Martin Sleziak Mar 8 '11 at 11:25
This way you can get only separable spaces with finite dimension and I guess you can get all of them (?) –  Anton Petrunin Mar 8 '11 at 15:48
@Anton Petrunin: Finite dimension can't be right, since every compact metrizable space is a quotient of the Cantor set, and that includes things like $[0,1]^{\aleph_0}$. –  Stephen S Mar 8 '11 at 21:17
Concerning your friends question: Take a look at Bill Lawvere's answer in mathoverflow.net/questions/127841/… "My paper about Volterra's functionals [...] discusses the unsuitability for functional analysis (as well as for homotopy theory) of the attempt to characterize continuity or cohesion using open sets or other contravariant structure." –  Garlef Wegart Feb 4 '14 at 17:38

Let $C\subset [0,1]$ denote the Cantor set, and for $k\in\mathbb{Z}$ let $C+k\subset[k,k+1]$ denote the shifted Cantor set as a subset of $\mathbb{R}$. If $X$ is the union of dyadic spaces $X_1$, $X_2$, ..., then $X$ is a continuous image of the space $\bigcup_k (C+2k)$, which is a closed subset of $\mathbb{R}$.
Conversely, assume $X$ is a quotient of a closed subset $Y\subset\mathbb{R}^n$. Then $Y$ is a countable union of compact subsets $Y_1$, $Y_2$, ... of $\mathbb{R}^n$. Since each $Y_k$ is compact and metrizable, its image in $X$ is dyadic.
This also shows that one obtains the same class by considering just quotients of closed subsets of $\mathbb{R}$.